準備

Tr[AB]=AijBji=BjiAij=Tr[BA]\begin{aligned} \text{Tr}[AB] &= A_{ij}B_{ji} \\ &= B_{ji} A_{i j} \\ &= \text{Tr}[BA] \end{aligned}

Tr[AB]=dψψABψ=dψψA(dψψψ)Bψ=dψdψψAψψBψ=dψdψψBψψAψ=dψψB(dψψψ)Aψ=dψψBAψ=Tr[BA]\begin{aligned} \text{Tr} [AB] &= \int d\psi \bra{\psi} AB \ket{\psi} \\ &= \int d\psi \bra{\psi} A \left( \int d\psi' \ket{\psi'}\bra{\psi'} \right) B \ket{\psi} \\ &= \int d\psi \int d\psi' \bra{\psi} A \ket{\psi'}\bra{\psi'} B \ket{\psi} \\ &= \int d\psi' \int d\psi \bra{\psi'} B \ket{\psi} \bra{\psi} A \ket{\psi'} \\ &= \int d\psi' \bra{\psi'} B \left( \int d\psi \ket{\psi}\bra{\psi} \right) A \ket{\psi'} \\ &= \int d\psi' \bra{\psi'} BA \ket{\psi'} \\ &= \text{Tr} [BA] \end{aligned}

ABeq=ψψABπψ=Tr[ABπ]=Tr[πAB]     (トレースの循環性)=Tr[BπA]     (トレースの循環性)Tr[πBA]     (ie.BAeq)\begin{aligned} \langle A B \rangle_{eq} &= \sum_{\psi} \bra{\psi} A B \pi \ket{\psi} \\ &= \text{Tr}\left[ A B \pi \right] \\ &= \text{Tr}\left[ \pi A B \right] \ \ \ \ \ (\because \text{トレースの循環性}) \\ &= \text{Tr}\left[ B \pi A \right] \ \ \ \ \ (\because \text{トレースの循環性}) \\ &\neq \text{Tr}\left[ \pi B A \right] \ \ \ \ \ (ie. \neq \langle B A \rangle_{eq}) \\ \end{aligned}

U(t)=eihHtU(t)=U(t)1=eihHt\begin{align*} &U_{(t)} = e^{-\frac{i}{h}Ht} \\ &U^{\dagger}_{(t)} = U^{-1}_{(t)} = e^{\frac{i}{h}Ht} \\ \end{align*}

A(t)=U(t)AU(t)A˙(t)=ddt(U(t)AU(t))=U(t)iHAU(t)U(t)AiHU(t)=U(t)i[H,A]U(t)=U(t)A˙U(t)\begin{aligned} A_{(t)} &= U^{\dagger}_{(t)} A U_{(t)} \\ \dot{A}_{(t)} &= \frac{d}{dt} \left( U^{\dagger}_{(t)} A U_{(t)} \right) \\ &= U^{\dagger}_{(t)} \frac{i}{\hbar} H A U_{(t)} - U^{\dagger}_{(t)} A \frac{i}{\hbar} H U_{(t)} \\ &= U^{\dagger}_{(t)} \frac{i}{\hbar} [H, A] U_{(t)} \\ &= U^{\dagger}_{(t)} \dot{A} U_{(t)} \end{aligned}

itψ(t)=H(t)ψ(t)itU(t)=H(t)U(t)U(t+Δt)=(IiH(t)Δt)U(t)\begin{aligned} i \hbar \frac{\partial}{\partial t} \ket{\psi_{(t)}} &= H_{(t)} \ket{\psi_{(t)}} \\ \therefore i \hbar \frac{\partial}{\partial t} U_{(t)} &= H_{(t)} U_{(t)} \\ \therefore U_{(t + \Delta t)} &= \left( I - \frac{i}{\hbar} H_{(t)} \Delta t \right) U_{(t)} \end{aligned}

逐次積分

U(Δt)=(IiH(0)Δt)U(0)=(IiH(0)Δt)U(2Δt)=(IiH(Δt)Δt)U(Δt)=(IiH(Δt)Δt)(IiH(0)Δt)U(nΔt)=in(IiH(iΔt)Δt)U(τ)=limnin(IiH(iτn)τn)=Ii0τdtH(t)+(i)2ττdt0τdtH(dt)H(dt)+exp(i0τHdt)     (時間順序指数)\begin{aligned} U_{(\Delta t)} &= \left(I - \frac{i}{\hbar}H_{(0)} \Delta t \right) U_{(0)} = \left( I - \frac{i}{\hbar} H_{(0)} \Delta t \right) \\ U_{(2\Delta t)} &= \left(I - \frac{i}{\hbar}H_{(\Delta t)} \Delta t \right) U_{(\Delta t)} = \left( I - \frac{i}{\hbar} H_{(\Delta t)} \Delta t \right) \left( I - \frac{i}{\hbar} H_{(0)} \Delta t \right) \\ \cdots \\ U_{(n\Delta t)} &= \prod^{n}_{i} \left( I - \frac{i}{\hbar} H_{(i \Delta t)} \Delta t \right) \\ \therefore U_{(\tau)} &= \lim_{n \to \infty} \prod^{n}_{i} \left( I - \frac{i}{\hbar} H_{(\frac{i\tau}{n})} \frac{\tau}{n} \right) = I - \frac{i}{\hbar} \int^{\tau}_0 dt H_{(t)} + \left( \frac{i}{\hbar} \right)^2 \int^{\tau}_{\tau'} dt \int^{\tau'}_0 dt' H_{(dt)} H_{(dt')} + \cdots \\ &\coloneqq \exp_{\leftarrow} \left( -\frac{i}{\hbar} \int^{\tau}_0 H dt \right) \ \ \ \ \ (\text{時間順序指数}) \end{aligned}

古典系

A(t)eqeL0tAeq=dΓ(eL0tA)π=dΓAeL0tπ=dΓAπ(t)=dΓAπ=Aeq\begin{aligned} \langle A_{(t)} \rangle_{eq} &\coloneqq \langle e^{L^{\dagger}_0 t} A \rangle_{eq} \\ &= \int d\Gamma (e^{L^{\dagger}_0 t} A) \pi \\ &= \int d\Gamma A e^{L_0 t} \pi \\ &= \int d\Gamma A \pi_{(-t)} \\ &= \int d\Gamma A \pi \\ &= \langle A \rangle_{eq} \end{aligned}

量子系

P1APeq=Tr[P1AP]=Tr[APP1]=Tr[A]=Aeq\begin{aligned} \langle P^{-1} A P \rangle_{eq} &= \text{Tr} [P^{-1} A P] \\ &= \text{Tr} [A P P^{-1}] \\ &= \text{Tr}[A] \\ &= \langle A \rangle_{eq} \end{aligned}

A(t)eq=UtAUteq=Aeq     (Utがユニタリ)\langle A_{(t)} \rangle_{eq} = \langle U^{\dagger}_t A U_t \rangle_{eq} = \langle A \rangle_{eq} \ \ \ \ \ (\because U_t\text{がユニタリ})

A(t)B(τ)eq=UtAUtUτBUτeq=AUtUτB(τt)UτUteq=AU(τt)B(τt)U(τt)eq=AB(τt)eq\begin{aligned} \langle A_{(t)} B_{(\tau)} \rangle_{eq} &= \langle U^{\dagger}_t A U_t U^{\dagger}_\tau B U_\tau \rangle_{eq} \\ &= \langle A U_t U^{\dagger}_\tau B_{(\tau - t)} U_\tau U^\dagger_t \rangle_{eq} \\ &= \langle A U^{\dagger}_{(\tau - t)} B_{(\tau - t)} U_{(\tau - t)} \rangle_{eq} \\ &= \langle A B_{(\tau - t)} \rangle_{eq} \end{aligned}

B˙(τ)A(t)eq=(ddτB(τ))A(t)eq=ddτB(τ)A(t)eq=ddτBA(tτ)eq=ddtBA(tτ)eq=BA˙(tτ)eq\begin{aligned} \left\langle \dot{B}_{(\tau)} A_{(t)} \right\rangle_{eq} &= \left\langle \left( \frac{d}{d\tau} B_{(\tau)} \right) A_{(t)} \right\rangle_{eq} \\ &= \frac{d}{d\tau} \left\langle B_{(\tau)} A_{(t)} \right\rangle_{eq} \\ &= \frac{d}{d\tau} \left\langle B A_{(t - \tau)} \right\rangle_{eq} \\ &= - \frac{d}{dt} \left\langle B A_{(t - \tau)} \right\rangle_{eq} \\ &= - \left\langle B \dot{A}_{(t - \tau)} \right\rangle_{eq} \\ \end{aligned}

2.2 古典粒子系に対する揺動散逸定理 (続き)

応答関数を計算するときは、系が平衡状態(H=Htot)(H = H_{tot})のときを考えればいよい。

α+O(ϵ)ϕx˙x(t;HtotϵxF(t))α=ϕx˙x(t;HtotϵxF(t))ϵ=0=ϕx˙x(t;Htot)\begin{aligned} \alpha + \text{O}_{(\epsilon)} \coloneqq \phi_{\dot{x}x(t; H_{tot} - \epsilon x F_{(t)})} \\ \alpha = \left. \phi_{\dot{x}x(t; H_{tot} - \epsilon x F_{(t)})} \right|_{\epsilon = 0} = \phi_{\dot{x}x(t; H_{tot})} \end{aligned}

x˙(τ)ϵτdtF(t)ϕx˙x(τt;HtotϵxF(t))=ϵτdtF(t)(α+O(ϵ))ϵτdtF(t)α=ϵτdtF(t)ϕx˙x(τt;Htot)\begin{aligned} \langle \dot{x}_{(\tau)} \rangle &\simeq \epsilon \int^{\tau}_{-\infty} dt F_{(t)} \phi_{\dot{x}x(\tau - t; H_{tot} - \epsilon x F_{(t)})} \\ &= \epsilon \int^{\tau}_{-\infty} dt F_{(t)} \left( \alpha + \text{O}_{(\epsilon)} \right) \\ &\simeq \epsilon \int^{\tau}_{-\infty} dt F_{(t)} \alpha \\ &= \epsilon \int^{\tau}_{-\infty} dt F_{(t)} \phi_{\dot{x}x(\tau - t; H_{tot})} \end{aligned}

応答関数 ϕx˙x\phi_{\dot{x}x}

x˙(t)=eL0tx˙=eL0tL0x=ddteL0tx=ddtx(t)\begin{aligned} \dot{x}_{(t)} &= e^{L^{\dagger}_0 t} \dot{x} \\ &= e^{L^{\dagger}_0 t} L^{\dagger}_0 x \\ &= \frac{d}{dt} e^{L^{\dagger}_0 t} x \\ &= \frac{d}{dt} x_{(t)} \end{aligned}

ϕx˙x(τt)βx˙(τt)x˙eq=βx˙(τ)x˙(t)eq=βx˙(τ)ddtx(t)eq=βddtx˙(τ)x(t)eq\begin{aligned} \phi_{\dot{x}x(\tau - t)} &\coloneqq \beta \langle \dot{x}_{(\tau - t)} \dot{x} \rangle_{eq} \\ &= \beta \langle \dot{x}_{(\tau)} \dot{x}_{(t)} \rangle_{eq} \\ &= \beta \langle \dot{x}_{(\tau)} \frac{d}{dt} x_{(t)} \rangle_{eq} \\ &= \beta \frac{d}{dt} \langle \dot{x}_{(\tau)} x_{(t)} \rangle_{eq} \end{aligned}

長時間後に一定になった速度 (2.11)\cdots (2.11)

x˙(τ)F0=τdtϕx˙x(τt)=0dtϕx˙x(t)     (tτt)=β0dtx˙(t)x˙eqx˙(τ)eqF0=x˙eqF0=β0dtx˙(t)x˙eq\begin{aligned} \frac{\langle \dot{x}_{(\tau)} \rangle}{F_0} &= \int^{\tau}_{-\infty} dt \phi_{\dot{x}x(\tau - t)} \\ &= \int^{\infty}_0 dt \phi_{\dot{x}x(t)} \ \ \ \ \ (t \Rightarrow \tau - t) \\ &= \beta \int^{\infty}_0 dt \langle \dot{x}_{(t)} \dot{x} \rangle_{eq} \\ \therefore \frac{\langle \dot{x}_{(\tau)} \rangle_{eq}}{F_0} &= \frac{\langle \dot{x} \rangle_{eq}}{F_0} = \beta \int^{\infty}_0 dt \langle \dot{x}_{(t)} \dot{x} \rangle_{eq} \end{aligned}

ランジュバン方程式でのμ\mu

mx¨=Fγx˙+ξm \ddot{x} = F - \gamma \dot{x} + \xi

mx¨=Fγx˙+ξF=γx˙     (速度一定)x˙F=μ=12kBTdsx˙(s)x˙     (ref. (1.12))\begin{aligned} m \langle \ddot{x} \rangle &= F - \gamma \langle \dot{x} \rangle + \langle \xi \rangle \\ \therefore F &= \gamma \langle \dot{x} \rangle \ \ \ \ \ (\because \text{速度一定}) \\ \therefore \frac{\langle \dot{x} \rangle}{F} &= \mu = \frac{1}{2k_B T} \int^{\infty}_{-\infty} ds \langle \dot{x}_{(s)} \dot{x} \rangle \ \ \ \ \ (\text{ref. } (1.12)) \end{aligned}

パルス入力に対する応答 (2.12)\cdots (2.12)

F(t)=F0δ(tt)F_{(t)} = F_0 \delta_{(t - t')}

x˙(τ)=ϵτdtF(t)ϕxx˙(τt)=ϵτdtF0δ(tt)ϕxx˙(τt)=ϵF(0)ϕxx˙(τt)     (τt)\begin{aligned} \langle \dot{x}_{(\tau)} \rangle &= \epsilon \int^{\tau}_{-\infty} dt F_{(t)} \phi_{x\dot{x}(\tau - t)} \\ &= \epsilon \int^{\tau}_{-\infty} dt F_0 \delta_{(t - t')} \phi_{x\dot{x}(\tau - t)} \\ &= \epsilon F_{(0)} \phi_{x\dot{x}(\tau - t')} \ \ \ \ \ (\tau \ge t') \end{aligned}

フーリエ表示の揺動散逸定理 (2.15)\cdots (2.15)

χx˙x(ω)=χ(ω)+iχ(ω)\chi_{\dot{x}x(\omega)} = \chi'_{(\omega)} + i \chi''_{(\omega)}

χ(ω)=Re[0dtβx˙(t)x˙eqeiωt]=β0dtC(t)Re[eiωt]=β0dtC(t)cos(ωt)=β20dtC(t)(eiωt+eiωt)=β2(0dtC(t)eiωt0(dt)C(t)eiωt)=β2(0dtC(t)eiωt+0dtC(t)eiωt)=β2C(ω)\begin{aligned} \chi'_{(\omega)} &= \text{Re} \left[ \int^{\infty}_{0} dt \beta \langle \dot{x}_{(t)} \dot{x} \rangle_{eq} e^{i\omega t} \right] \\ &= \beta \int^{\infty}_{0} dt C_{(t)} \text{Re} [e^{i\omega t}] \\ &= \beta \int^{\infty}_0 dt C_{(t)} \cos{(\omega t)} \\ &= \frac{\beta}{2} \int^{\infty}_{0} dt C_{(t)} \left( e^{i \omega t} + e^{-i \omega t} \right) \\ &= \frac{\beta}{2} \left( \int^{\infty}_{0} dt C_{(t)} e^{i \omega t} - \int^{\infty}_0 (-dt) C_{(-t)} e^{-i \omega t} \right) \\ &= \frac{\beta}{2} \left( \int^{\infty}_{0} dt C_{(t)} e^{i \omega t} + \int^0_{-\infty} dt C_{(t)} e^{i \omega t} \right) \\ &= \frac{\beta}{2} C_{(\omega)} \end{aligned}

2.2 線形応答理論

§ 2.2.1 力学応答の線形応答

Htot(t)=HBF(t)H_{tot(t)} = H - BF_{(t)}

タイプ 1

ABA \propto B

タイプ 2

AB˙A \propto \dot{B}

§ 2.2.1.1 一般論

時間発展演算子 (2.18)\cdots (2.18)

itU(t)=[HFB]U(t)Uˉ(t)eiHtU(t)\begin{aligned} i \hbar \frac{\partial}{\partial t} U_{(t)} &= [H - FB] U_{(t)} \\ \bar{U}_{(t)} &\coloneqq e^{\frac{i}{\hbar}Ht} U_{(t)} \\ \end{aligned}

tUˉ(t)=iHUˉ(t)ieiHt[HF(t)B]U(t)=ieiHtF(t)BU(t)=iF(t)(eiHtBeiHt)eiHtU(t)=F(t)B(t)Uˉ(t)\begin{align*} \frac{\partial}{\partial t} \bar{U}_{(t)} &= \frac{i}{\hbar} H \bar{U}_{(t)} - \frac{i}{\hbar} e^{\frac{i}{\hbar}Ht}[H - F_{(t)}B]U_{(t)} \\ &= \frac{i}{\hbar} e^{\frac{i}{\hbar}Ht} F_{(t)}B U_{(t)} \\ &= \frac{i}{\hbar} F_{(t)} \left( e^{\frac{i}{\hbar}Ht} B e^{-\frac{i}{\hbar}Ht} \right) e^{\frac{i}{\hbar} Ht} U_{(t)} \\ &= F_{(t)}B_{(t)} \bar{U}_{(t)} \\ \end{align*}

[note] FFについて一次の近似を考えるので、B の時間発展を考える時は、FFを考慮せず平衡状態のハミルトニアンで時間発展すると考えて良い。

Uˉ(t+Δt)=(I+iF(t)B(t)Δt)Uˉ(t)U(t+Δt)=eiH(t+Δt)(I+iF(t)B(t)Δt)eiHtU(t)U(t)=eiHtin(eiHΔt(I+iF(iΔt)B(iΔt)Δt))eiHt0eiHtexp(it0tdtF(t)B(t))eiHt0     (Δt0)\begin{aligned} \bar{U}_{(t + \Delta t)} &= \left( I + \frac{i}{\hbar} F_{(t)}B_{(t)} \Delta t \right) \bar{U}_{(t)} \\ \therefore U_{(t + \Delta t)} &= e^{- \frac{i}{\hbar}H(t+\Delta t)} \left( I + \frac{i}{\hbar} F_{(t)}B_{(t)} \Delta t \right) e^{\frac{i}{\hbar}Ht} U_{(t)} \\ \therefore U_{(t)} &= e^{- \frac{i}{\hbar} H t} \prod^{n}_{i} \left( e^{- \frac{i}{\hbar} H \Delta t} \left( I + \frac{i}{\hbar} F_{(i\Delta t)}B_{(i \Delta t)} \Delta t \right) \right) e^{\frac{i}{\hbar}Ht_0} \\ &\to e^{- \frac{i}{\hbar} Ht} \exp_{\leftarrow} \left( \frac{i}{\hbar} \int^t_{t_0} dt' F_{(t')} B_{(t')} \right) e^{\frac{i}{\hbar}Ht_0} \ \ \ \ \ (\Delta t \to 0) \end{aligned}

U(t)=eiHtin(eiHΔt(I+iF(iΔt)B(iΔt)Δt))eiHt0=eiHt(eiHΔt)n[1+in(iF(iΔt)B(iΔt)Δt)]eiHt0+O(F2)eiHt(1+it0tdtF(t)B(t))eiHt0+O(F2)\begin{aligned} U_{(t)} &= e^{- \frac{i}{\hbar} H t} \prod^{n}_{i} \left( e^{- \frac{i}{\hbar} H \Delta t} \left( I + \frac{i}{\hbar} F_{(i\Delta t)}B_{(i \Delta t)} \Delta t \right) \right) e^{\frac{i}{\hbar}Ht_0} \\ &= e^{- \frac{i}{\hbar} H t} \left( e^{- \frac{i}{\hbar} H \Delta t} \right)^n \left[ 1 + \sum^{n}_{i} \left( \frac{i}{\hbar} F_{(i\Delta t)}B_{(i \Delta t)} \Delta t \right) \right] e^{\frac{i}{\hbar}Ht_0} + \text{O}(F^2) \\ &\to e^{- \frac{i}{\hbar} H t} \left( 1 + \frac{i}{\hbar} \int^t_{t_0} dt' F_{(t')} B_{(t')} \right) e^{\frac{i}{\hbar} Ht_0} + \text{O}(F^2) \end{aligned}

演算子の期待値の時間発展 (2.19)\cdots (2.19)

A(τ)=U(τ)AU(τ)=eiHt0(1it0τdtF(t)B(t))eiHτAeiHτ(1+it0τdtF(t)B(t))eiHt0+O(F2)=eiH(τt0)AeiH(τt0)+eiHt0[eiHτAeiHτ,it0τdtF(t)B(t)]eiHt0+O(F2)=eiH(τt0)AeiH(τt0)+it0τdtF(t)[A(τt0),B(tt0)]+O(F2)\begin{aligned} A_{(\tau)} &= U^{\dagger}_{(\tau)} A U_{(\tau)} \\ &= e^{- \frac{i}{\hbar} H t_0} \left( 1 - \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')} B_{(t')} \right) e^{\frac{i}{\hbar} H \tau} A e^{- \frac{i}{\hbar} H \tau} \left( 1 + \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')} B_{(t')} \right) e^{\frac{i}{\hbar} Ht_0} + \text{O}(F^2) \\ &= e^{\frac{i}{\hbar} H (\tau - t_0)} A e^{-\frac{i}{\hbar} H (\tau - t_0)} + e^{-\frac{i}{\hbar}Ht_0} \left[ e^{\frac{i}{\hbar} H \tau} A e^{- \frac{i}{\hbar} H \tau}, \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')}B_{(t')} \right] e^{\frac{i}{\hbar}Ht_0} + \text{O}(F^2) \\ &= e^{\frac{i}{\hbar} H (\tau - t_0)} A e^{-\frac{i}{\hbar} H (\tau - t_0)} + \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')} [ A_{(\tau - t_0)}, B_{(t' - t_0)} ] + \text{O}(F^2) \\ \end{aligned}

[note] FFの一次近似を考えているので、積分の中では非平衡の時間発展と平衡の時間発展を区別する必要がない。

A(τ)eqAeq=eiH(τt0)AeiH(τt0)eq+it0τdtF(t)[A(τt0),B(tt0)]eqAeq+O(F2)=Aeq+it0τdtF(t)[A(τt),B]eqAeq+O(F2)=it0τdtF(t)[A(τt),B]eq+O(F2)\begin{aligned} \langle A_{(\tau)} \rangle_{eq} - \langle A \rangle_{eq} &= \langle e^{\frac{i}{\hbar} H (\tau - t_0)} A e^{-\frac{i}{\hbar} H (\tau - t_0)} \rangle_{eq} + \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')} \langle [ A_{(\tau - t_0)}, B_{(t' - t_0)} ] \rangle_{eq} - \langle A \rangle_{eq} + \text{O}(F^2) \\ &= \langle A \rangle_{eq} + \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')} \langle [ A_{(\tau - t')}, B ] \rangle_{eq} - \langle A \rangle_{eq} + \text{O}(F^2) \\ &= \frac{i}{\hbar} \int^\tau_{t_0} dt' F_{(t')} \langle [ A_{(\tau - t')}, B ] \rangle_{eq} + \text{O}(F^2) \\ \end{aligned}

十分に時間が経った時の期待値の変化量 (2.20)\cdots (2.20)

ΔA(τ)eqlimt0A(τ)eqAeq=iτdtF(t)[A(τt),B]eq=i0dtF(τt)[A(t),B]eq     (tτt)=0dtϕAB(t)F(τt)\begin{aligned} \langle \Delta A_{(\tau)} \rangle_{eq} &\coloneqq \lim_{t_0 \to - \infty} \langle A_{(\tau)} \rangle_{eq} - \langle A \rangle_{eq} \\ &= \frac{i}{\hbar} \int^{\tau}_{- \infty} dt F_{(t)} \langle [ A_{(\tau - t)}, B] \rangle_{eq} \\ &= \frac{i}{\hbar} \int^{\infty}_{0} dt F_{(\tau - t)} \langle [ A_{(t)}, B] \rangle_{eq} \ \ \ \ \ (t \to \tau - t) \\ &= \int^{\infty}_{0} dt \phi_{AB(t)} F_{(\tau - t)} \end{aligned}

量子系の応答関数 (2.21)\cdots (2.21)

ϕAB(t)i[A(t),B]eq\phi_{AB(t)} \coloneqq \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq}

ΔA(τ)eq=F0ϕAB(τ)     (when. F=F0δ(t))\begin{aligned} \langle \Delta A_{(\tau)} \rangle_{eq} = F_0 \phi_{AB(\tau)} \ \ \ \ \ (\text{when. }F = F_0 \delta_{(t)}) \\ \end{aligned}

ϕAB(t)=i[A(t),B]eq=iTr[π[A(t),B]]=i(Tr[πA(t)B]Tr[πBA(t)])=1i(Tr[A(t)Bπ]+Tr[A(t)πB])=1iTr[A(t)[π,B]]\begin{aligned} \phi_{AB(t)} &= \frac{i}{\hbar}\langle [A_{(t)}, B] \rangle_{eq} \\ &= \frac{i}{\hbar} \text{Tr} \left[ \pi [A_{(t)}, B] \right] \\ &= \frac{i}{\hbar} \left( \text{Tr} \left[ \pi A_{(t)} B \right] - \text{Tr} \left[ \pi B A_{(t)} \right] \right) \\ &= - \frac{1}{i\hbar} \left( - \text{Tr} \left[ A_{(t)} B \pi \right] + \text{Tr} \left[ A_{(t)} \pi B \right] \right) \\ &= \frac{1}{i\hbar} \text{Tr} \left[ A_{(t)} [\pi, B] \right] \end{aligned}

カノニカル相関 (2.22)(2.27)\cdots (2.22) \sim (2.27)

X;Y1β0βduTr[euHXeuHYπ]=1β0βdu(Tr[euHXeuHYπ])=1β0βduTr[πYeuHXeuH]=1β0βduTr[euHπYeuHX]=1β0βduTr[πeuHYeuHX]=Y;X\begin{aligned} \langle X; Y \rangle &\coloneqq \frac{1}{\beta} \int^{\beta}_0 du \text{Tr} \left[ e^{uH} X e^{-u H} Y \pi \right] \\ &= \frac{1}{\beta} \int^{\beta}_0 du \left( \text{Tr} \left[ e^{uH} X e^{-u H} Y \pi \right] \right)^{\dagger} \\ &= \frac{1}{\beta} \int^{\beta}_0 du \text{Tr} \left[ \pi Y e^{-u H} X e^{uH} \right] \\ &= \frac{1}{\beta} \int^{\beta}_0 du \text{Tr} \left[ e^{uH} \pi Y e^{-u H} X \right] \\ &= \frac{1}{\beta} \int^{\beta}_0 du \text{Tr} \left[ \pi e^{uH} Y e^{-u H} X \right] \\ &= \langle Y; X \rangle \end{aligned}

X;Y=1β0βduX(iu)YeqXYeq     (0)\langle X; Y \rangle = \frac{1}{\beta} \int^{\beta}_0 du \langle X_{(-i\hbar u)} Y \rangle_{eq} \to \langle X Y \rangle_{eq} \ \ \ \ \ (\hbar \to 0)

[π,B]=1Z[eβH,B]=1Z(eβHBBeβH)=1Z(eβHBeβHB)eβH=1Z[euHBeuH]u=0u=βeβH=1Z0βduddu(euHBeuH)eβH=iZ0βdudd(iu)(eiH(iu)BeiH(iu))eβH=iZ0βduB˙(iu)eβH=i0βduB˙(iu)π[π,B]=1ZeβH(BeβHBeβH)==iπ0βduB˙(iu)\begin{aligned} [\pi, B] &= \frac{1}{Z} [e^{-\beta H}, B] \\ &= \frac{1}{Z} (e^{-\beta H} B - B e^{-\beta H}) \\ &= \frac{1}{Z} (e^{-\beta H} B e^{\beta H} - B) e^{-\beta H} \\ &= \frac{1}{Z} \left[ e^{-u H} B e^{u H} \right]^{u = \beta}_{u = 0} e^{-\beta H} \\ &= \frac{1}{Z} \int^{\beta}_0 du \frac{d}{du} (e^{-u H} B e^{u H}) e^{-\beta H} \\ &= \frac{i\hbar}{Z} \int^{\beta}_0 du \frac{d}{d(i \hbar u)} (e^{\frac{i}{\hbar} H (i \hbar u)} B e^{- \frac{i}{\hbar} H (i \hbar u)}) e^{-\beta H} \\ &= \frac{i\hbar}{Z} \int^{\beta}_0 du \dot{B}_{(i\hbar u)} e^{-\beta H} \\ &= i\hbar \int^{\beta}_0 du \dot{B}_{(i\hbar u)} \pi \\ \\ [\pi, B] &= \frac{1}{Z} e^{-\beta H} (B - e^{\beta H} B e^{-\beta H}) \\ &= \cdots \\ &= i\hbar \pi \int^{\beta}_0 du \dot{B}_{(-i\hbar u)} \end{aligned}

ϕAB(t)=1iTr[A(t)[π,B]]=1iTr[A(t)i0βduB˙(iu)π]\begin{aligned} \phi_{AB(t)} &= \frac{1}{i\hbar} \text{Tr} \left[A_{(t)} [\pi, B] \right] \\ &= \frac{1}{i\hbar} \text{Tr} \left[ A_{(t)} i\hbar \int^{\beta}_0 du \dot{B}_{(i\hbar u)} \pi \right] \\ \end{aligned}

ϕAB(t)=1iTr[A(t)[π,B]]=1iTr[A(t)i0βduB˙(iu)π]=0βduTr[A(t)B˙(iu)π]=0βduA(t)B˙(iu)eq=0βduA(tiu)B˙eq=βA(t);B˙=βB˙;A(t)\begin{aligned} \phi_{AB(t)} &= \frac{1}{i\hbar} \text{Tr} \left[A_{(t)} [\pi, B] \right] \\ &= \frac{1}{i\hbar} \text{Tr} \left[ A_{(t)} i\hbar \int^{\beta}_0 du \dot{B}_{(i\hbar u)} \pi \right] \\ &= \int^\beta_0 du \text{Tr} \left[ A_{(t)} \dot{B}_{(i\hbar u)} \pi \right] \\ &= \int^\beta_0 du \langle A_{(t)} \dot{B}_{(i\hbar u)} \rangle_{eq} \\ &= \int^\beta_0 du \langle A_{(t - i\hbar u)} \dot{B} \rangle_{eq} \\ &= \beta \langle A_{(t)}; \dot{B} \rangle \\ &= \beta \langle \dot{B}; A_{(t)} \rangle \end{aligned}

ϕAB(t)βA(t)B˙eq     (0)\phi_{AB(t)} \to \beta \langle A_{(t)} \dot{B} \rangle_{eq} \ \ \ \ \ (\hbar \to 0)

複素アドミッタンス (2.28)\cdots (2.28)

χAB(ω)0dtϕAB(t)eiωt=χ+iχ\chi_{AB(\omega)} \coloneqq \int^{\infty}_0 dt \phi_{AB(t)} e^{i\omega t} = \chi' + i \chi''

χAB(ω)=0dtϕAB(t)eiωt=0dtϕAB(t)(eiωt)=(0dtϕAB(t)eiωt)=χAB(ω)\begin{aligned} \chi_{AB(-\omega)} &= \int^{\infty}_0 dt \phi_{AB(t)} e^{-i \omega t} \\ &= \int^{\infty}_0 dt \phi_{AB(t)} \left( e^{i \omega t} \right)^{*} \\ &= \left( \int^{\infty}_0 dt \phi_{AB(t)} e^{i \omega t} \right)^{*} \\ &= \chi_{AB(\omega)}^{*} \end{aligned}

演算子の応答 (2.29)\cdots (2.29)

F(t)=F0cos(ωt)F_{(t)} = F_0 \cos{(\omega t)}

ΔA(τ)eq=0dtϕAB(t)F(τt)=0dtϕAB(t)F0cosω(τt)=F00dtϕAB(t)Re[eiω(τt)]=F0Re[0dtϕAB(t)eiω(τt)]=F0Re[0dtϕAB(t)eiωteiωτ]=F0Re[χAB(ω)eiωτ]=F0Re[χAB(ω)eiωτ]\begin{aligned} \langle \Delta A_{(\tau)} \rangle_{eq} &= \int^{\infty}_{0} dt \phi_{AB(t)} F_{(\tau - t)} \\ &= \int^{\infty}_{0} dt \phi_{AB(t)} F_0 \cos{\omega (\tau - t)} \\ &= F_0 \int^{\infty}_{0} dt \phi_{AB(t)} \text{Re} [e^{i \omega (\tau - t)}] \\ &= F_0 \text{Re} \left[ \int^{\infty}_{0} dt \phi_{AB(t)} e^{i \omega (\tau - t)} \right] \\ &= F_0 \text{Re} \left[ \int^{\infty}_{0} dt \phi_{AB(t)} e^{- i \omega t} e^{i \omega \tau} \right] \\ &= F_0 \text{Re} \left[ \chi_{AB(\omega)}^{*} e^{i \omega \tau} \right] \\ &= F_0 \text{Re} \left[ \chi_{AB(\omega)} e^{-i \omega \tau} \right] \\ \end{aligned}

§ 2.2.1.2 クラマース・クローニッヒ関係

リーマン・ルベーグの補題

limω00dtϕAB(t)eiωt=0     (Im[ω]>0)\lim_{\omega \to 0} \left| \int^{\infty}_0 dt \phi_{AB(t)} e^{i \omega t} \right| = 0 \ \ \ \ \ (\text{Im}[\omega] > 0)

I0dtϕAB(t)eiωt=πωdsϕAB(s+πω)eiω(s+πω)     (s=tπω)=πωdsϕAB(s+πω)eiωseiπ=(0dsϕAB(s+πω)eiωs+πω0dsϕAB(s+πω)eiωs)2I=0dt(ϕAB(t)ϕAB(t+πω))eiωtπω0dtϕAB(t+πω)eiωt2I0dt(ϕAB(t)ϕAB(t+πω))eiωt+πω0dtϕAB(t+πω)eiωt0dtϕAB(t)ϕAB(t+πω)eiωt+πω0dtϕAB(t+πω)eiωt<0dtϕAB(t)ϕAB(t+πω)+πω0dtϕAB(t+πω)     (Im[ω]>0)\begin{aligned} I &\coloneqq \int^{\infty}_0 dt \phi_{AB(t)} e^{i\omega t} \\ &= \int^{\infty}_{- \frac{\pi}{\omega}} ds \phi_{AB(s + \frac{\pi}{\omega})} e^{i \omega (s + \frac{\pi}{\omega})} \ \ \ \ \ (s = t - \frac{\pi}{\omega}) \\ &= \int^{\infty}_{- \frac{\pi}{\omega}} ds \phi_{AB(s + \frac{\pi}{\omega})} e^{i \omega s} e^{i \pi} \\ &= - \left( \int^{\infty}_0 ds \phi_{AB(s + \frac{\pi}{\omega})} e^{i \omega s} + \int^0_{- \frac{\pi}{\omega}} ds \phi_{AB(s + \frac{\pi}{\omega})} e^{i \omega s} \right) \\ \therefore 2I &= \int^{\infty}_0 dt \left( \phi_{AB(t)} - \phi_{AB(t + \frac{\pi}{\omega})} \right) e^{i \omega t} - \int^{0}_{- \frac{\pi}{\omega}} dt \phi_{AB(t + \frac{\pi}{\omega})} e^{i \omega t} \\ \therefore \left| 2I \right| &\le \left| \int^{\infty}_0 dt \left( \phi_{AB(t)} - \phi_{AB(t + \frac{\pi}{\omega})} \right) e^{i \omega t} \right| + \left| \int^{0}_{- \frac{\pi}{\omega}} dt \phi_{AB(t + \frac{\pi}{\omega})} e^{i \omega t} \right| \\ &\le \int^{\infty}_0 dt \left| \phi_{AB(t)} - \phi_{AB(t + \frac{\pi}{\omega})} \right| \left| e^{i \omega t} \right| + \int^{0}_{- \frac{\pi}{\omega}} dt \left| \phi_{AB(t + \frac{\pi}{\omega})} \right| \left| e^{i \omega t} \right| \\ &\lt \int^{\infty}_0 dt \left| \phi_{AB(t)} - \phi_{AB(t + \frac{\pi}{\omega})} \right| + \int^{0}_{- \frac{\pi}{\omega}} dt \left| \phi_{AB(t + \frac{\pi}{\omega})} \right| \ \ \ \ \ (\because \text{Im}[\omega] > 0) \\ \end{aligned}

0dtϕAB(t)ϕAB(t+πω)=limτ0τdtϕAB(t)ϕAB(t+πω)limτ0τdt(ϕAB(t)+ϕAB(t+πω))=20dtϕAB(t)     (有界かつτについて単調増加)\begin{aligned} \int^{\infty}_0 dt \left| \phi_{AB(t)} - \phi_{AB(t + \frac{\pi}{\omega})} \right| &= \lim_{\tau \to \infty} \int^{\tau}_0 dt \left| \phi_{AB(t)} - \phi_{AB(t + \frac{\pi}{\omega})} \right| \\ &\le \lim_{\tau \to \infty} \int^{\tau}_0 dt \left( \left| \phi_{AB(t)} \right| + \left| \phi_{AB(t + \frac{\pi}{\omega})} \right| \right) \\ &= 2 \int^{\infty}_0 dt \left| \phi_{AB(t)} \right| \ \ \ \ \ (\Rightarrow \text{有界かつ}\tau\text{について単調増加}) \end{aligned}

limωI=limΔt0120dtϕAB(t+Δt)ϕAB(t)=limΔt012limΔt0iϕAB(ti+Δt)ϕAB(ti)Δt=limΔt012iϕAB(ti+1)ϕAB(ti)Δt     (Δt=Δtの条件で極限をとる)limΔt012i(Mimi)Δt=limΔt012(S(Δt)s(Δt))=0     (ϕAB(t)はリーマン可積分)\begin{aligned} \therefore \lim_{\omega \to \infty} \left| I \right| &= \lim_{\Delta t \to 0} \frac{1}{2} \int^{\infty}_0 dt \left| \phi_{AB(t + \Delta t)} - \phi_{AB(t)} \right| \\ &= \lim_{\Delta t \to 0} \frac{1}{2} \lim_{\Delta t' \to 0} \sum_{i} \left| \phi_{AB(t_i + \Delta t)} - \phi_{AB(t_i)} \right| \Delta t' \\ &= \lim_{\Delta t \to 0} \frac{1}{2} \sum_{i} \left| \phi_{AB(t_{i+1})} - \phi_{AB(t_i)} \right| \Delta t \ \ \ \ \ (\Delta t' = \Delta t \text{の条件で極限をとる})\\ &\le \lim_{\Delta t \to 0} \frac{1}{2} \sum_{i} ( M_i - m_i ) \Delta t \\ &= \lim_{\Delta t \to 0} \frac{1}{2} \left( S_{(\Delta t)} - s_{(\Delta t)} \right) \\ &= 0 \ \ \ \ \ (\because \phi_{AB(t)}\text{はリーマン可積分}) \end{aligned}

複素アドミッタンス

χAB(ω)=0ϕAB(t)eiωt\begin{aligned} \chi_{AB(\omega)} &= \int^{\infty}_0 \phi_{AB(t)} e^{i\omega t} \end{aligned}

badxϵx2+ϵ2f(x)=12ibadx(1xiϵ1x+iϵ)f(x)=12ibiϵaiϵdzf(z+iϵ)z+a+iϵb+iϵdzf(ziϵ)z=12idzf(z)z+O(ϵ){πf(0)     (原点が積分経路に含まれる)0     (原点が積分経路に含まれない)\begin{aligned} \int^a_b dx \frac{\epsilon}{x^2 + \epsilon^2} f_{(x)} &= \frac{1}{2i} \int^a_b dx \left( \frac{1}{x - i\epsilon} - \frac{1}{x + i\epsilon} \right) f_{(x)} \\ &= \frac{1}{2i} \int^{a - i\epsilon}_{b - i\epsilon} dz \frac{f_{(z + i\epsilon)}}{z} + \int^{b + i\epsilon} _{a + i\epsilon} dz \frac{f_{(z - i\epsilon)}}{z} \\ &= \frac{1}{2i} \oint dz \frac{f_{(z)}}{z} + \text{O}_{(\epsilon)} \\ &\to \left\{ \begin{split} &\pi f_{(0)} \ \ \ \ \ (\text{原点が積分経路に含まれる}) \\ &0 \ \ \ \ \ (\text{原点が積分経路に含まれない}) \end{split} \right. \end{aligned}

limϵ0ϵx2+ϵ2=πδ(x)\therefore \lim_{\epsilon \to 0} \frac{\epsilon}{x^2 + \epsilon^2} = \pi \delta(x)

1xiϵ=x+iϵ(xiϵ)(x+iϵ)=xx2+ϵ2+iϵx2+ϵ2Pv1x+iπδ(x)\begin{aligned} \frac{1}{x - i\epsilon} &= \frac{x + i\epsilon}{(x - i\epsilon)(x + i\epsilon)} \\ &= \frac{x}{x^2 + \epsilon^2} + i \frac{\epsilon}{x^2 + \epsilon^2} \\ &\to \text{Pv} \frac{1}{x} + i \pi \delta_{(x)} \end{aligned}

複素平面に拡張された複素アドミッタンス

12πidωχAB(ω)ωz={χAB(z)     (Im[z]>0)0     (Im[z]<0)\frac{1}{2\pi i} \int^{\infty}_{- \infty} d\omega' \frac{\chi_{AB(\omega')}}{\omega' - z} = \left\{ \begin{aligned} &\chi_{AB(z)} \ \ \ \ \ (\text{Im}[z] > 0) \\ &0 \ \ \ \ \ (\text{Im}[z] < 0) \\ \end{aligned} \right.

12πidωχAB(ω)ω(ω+iϵ)\frac{1}{2\pi i} \int^{\infty}_{\infty} d\omega' \frac{\chi_{AB(\omega')}}{\omega' - (\omega + i \epsilon)}

12πidωχAB(ω)ω(ω+iϵ)=12πidωχAB(ω)(ωω)iϵ12πidω(PvχAB(ω)ωω+iπχAB(ω)δ(ωω))=12πiPvdωχAB(ω)ωω+12χAB(ω)\begin{aligned} \frac{1}{2\pi i} \int^{\infty}_{-\infty} d\omega' \frac{\chi_{AB(\omega')}}{\omega' - (\omega + i \epsilon)} &= \frac{1}{2\pi i} \int^{\infty}_{\infty} d\omega' \frac{\chi_{AB(\omega')}}{(\omega' - \omega) - i \epsilon} \\ &\to \frac{1}{2 \pi i} \int^{\infty}_{-\infty} d\omega' \left( \text{Pv} \frac{\chi_{AB(\omega')}}{\omega' - \omega} + i \pi \chi_{AB(\omega')} \delta_{(\omega' - \omega)} \right) \\ &= \frac{1}{2 \pi i} \text{Pv} \int^{\infty}_{-\infty} d\omega' \frac{\chi_{AB(\omega')}}{\omega' - \omega} + \frac{1}{2} \chi_{AB(\omega)} \\ \end{aligned}

クラマース・クローニッヒ関係 (2.30)\cdots (2.30)

χAB(ω)+iχAB(ω)=12πiPvdωχAB(ω)+iχAB(ω)ωω+12(χAB(ω)+iχAB(ω))χAB(ω)+iχAB(ω)=1πPvdωiχAB(ω)+χAB(ω)ωω\begin{aligned} \chi'_{AB(\omega)} + i\chi''_{AB(\omega)} &= \frac{1}{2 \pi i} \text{Pv} \int^{\infty}_{- \infty} d\omega' \frac{\chi'_{AB(\omega')} + i \chi''_{AB(\omega')}}{\omega' - \omega} + \frac{1}{2} \left( \chi'_{AB(\omega)} + i \chi''_{AB(\omega)} \right) \\ \therefore \chi'_{AB(\omega)} + i\chi''_{AB(\omega)} &= \frac{1}{\pi} \text{Pv} \int^{\infty}_{-\infty} d\omega' \frac{-i \chi'_{AB(\omega')} + \chi''_{AB(\omega')}}{\omega' - \omega} \end{aligned}

χAB(ω)=1πPvdωχAB(ω)ωωχAB(ω)=1πPvdωχAB(ω)ωω\begin{aligned} \chi'_{AB(\omega)} &= \frac{1}{\pi} \text{Pv} \int^{\infty}_{-\infty} d\omega' \frac{\chi''_{AB(\omega')}}{\omega' - \omega} \\ \chi''_{AB(\omega)} &= - \frac{1}{\pi} \text{Pv} \int^{\infty}_{-\infty} d\omega' \frac{\chi'_{AB(\omega')}}{\omega' - \omega} \\ \end{aligned}

2.2.2 力学応答の揺動散逸定理 (P. 27)

§ 2.2.2.1 パワーロス (P. 27)

Htot=HBF(t)H_{tot} = H - BF_{(t)}

タイプ 1

ΔA(t)=B(t)\Delta A_{(t)} = B_{(t)}

タイプ 2

ΔA(t)=B˙(t)\Delta A_{(t)} = \dot{B}_{(t)}

一周期あたりの仕事 W (2.31)\cdots(2.31)

W=02πωdtdF(t)dtTr(HtotFρ(t))=02πωdtdF(t)dtB(t)eq     (HtotF=B)\begin{aligned} W &= \int^{\frac{2\pi}{\omega}}_{0} dt \frac{dF_{(t)}}{dt} \text{Tr}\left( \frac{\partial H_{tot}} {\partial F} \rho_{(t)} \right) \\ &= - \int^{\frac{2\pi}{\omega}}_{0} dt \frac{dF_{(t)}}{dt} \langle B_{(t)} \rangle_{eq} \ \ \ \ \ \left( \because \frac{\partial H_{tot}} {\partial F} = -B \right) \\ \end{aligned}

単位時間あたりの仕事PP (2.32)\cdots(2.32)

P(1)=ω2π02πωdtdF(t)dtB(t)eq=ω2π02πωdtdF(t)dtΔA(t)eq     (ΔA(t)=B(t))=ω2π02πωdt(ω)F0sin(ωt)F0Re[χAB(ω)eiωt]=ω2F022πRe[χAB(ω)02πωdtsin(ωt)eiωt]=ω2F022πRe[χAB(ω)02πωdt(i)sin2(ωt)]=ωF022χAB(ω)\begin{aligned} P^{(1)} &= - \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt \frac{dF_{(t)}}{dt} \langle B_{(t)} \rangle_{eq} \\ &= - \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt \frac{dF_{(t)}}{dt} \langle \Delta A_{(t)} \rangle_{eq} \ \ \ \ \ \left(\because \Delta A_{(t)} = B_{(t)} \right) \\ &= - \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt (-\omega) F_0 \sin{(\omega t)} F_0 \text{Re} \left[ \chi_{AB(\omega)} e^{-i\omega t} \right] \\ &= \frac{\omega^2 F_0^2}{2 \pi} \text{Re} \left[ \chi_{AB(\omega)} \int^{\frac{2\pi}{\omega}}_{0} dt \sin{(\omega t)} e^{-i\omega t} \right] \\ &= \frac{\omega^2 F_0^2}{2 \pi} \text{Re} \left[ \chi_{AB(\omega)} \int^{\frac{2\pi}{\omega}}_{0} dt (-i) \sin^2{(\omega t)} \right] \\ &= \frac{\omega F_0^2}{2} \chi''_{AB(\omega)} \end{aligned}

P(2)=ω2π02πωdtdF(t)dtB(t)eq=ω2π[F(t)B(t)eq]02πω+ω2π02πωdtF(t)ddtB(t)eq=ω2π02πωdtF(t)B˙(t)eq     (定常状態の周期性)=ω2π02πωdtF(t)ΔA(t)eq     (ΔA(t)=B˙(t))=ω2π02πωdtF0cos(ωt)F0Re[χAB(ω)eiωt]=ω2πF02Re[χAB(ω)02πωdtcos(ωt)eiωt]=ω2πF02Re[χAB(ω)02πωdtcos2(ωt)]=F022χAB(ω)\begin{aligned} P^{(2)} &= - \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt \frac{dF_{(t)}}{dt} \langle B_{(t)} \rangle_{eq} \\ &= - \frac{\omega}{2 \pi} \left[ F_{(t)} \langle B_{(t)} \rangle_{eq} \right]^{\frac{2\pi}{\omega}}_{0} + \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt F_{(t)} \frac{d}{dt} \langle B_{(t)} \rangle_{eq} \\ &= \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt F_{(t)} \langle \dot{B}_{(t)} \rangle_{eq} \ \ \ \ \ \left( \because \text{定常状態の周期性} \right) \\ &= \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt F_{(t)} \langle \Delta A_{(t)} \rangle_{eq} \ \ \ \ \ \left( \because \Delta A_{(t)} = \dot{B}_{(t)} \right) \\ &= \frac{\omega}{2 \pi} \int^{\frac{2\pi}{\omega}}_{0} dt F_0 \cos{(\omega t)} F_0 \text{Re} \left[ \chi_{AB(\omega)} e^{-i\omega t} \right] \\ &= \frac{\omega}{2 \pi} F_{0}^{2} \text{Re} \left[ \chi_{AB(\omega)} \int^{\frac{2\pi}{\omega}}_{0} dt \cos{(\omega t)} e^{-i\omega t} \right] \\ &= \frac{\omega}{2 \pi} F_{0}^{2} \text{Re} \left[ \chi_{AB(\omega)} \int^{\frac{2\pi}{\omega}}_{0} dt \cos^2{(\omega t)} \right] \\ &= \frac{F_0^2}{2} \chi'_{AB(\omega)} \end{aligned}

等温過程応答係数χABT\chi^{T}_{AB} (2.34)\cdots(2.34)

χABT=F0Tr[eβ(HF0B)A]Tr[eβ(HF0B)]F00=1Z{F0Tr[eβ(HF0B)A]AeqF0Tr[eβ(HF0B)]}F00=1Z{Tr[01dyeβH(1y)(βB)eβHyA]AeqTr[01dyeβH(1y)(βB)eβHy]}=1ZTr[01dyeβH(1y)(βB)eβHy(AAeq)]=1ZTr[0βdue(uHβH)BeuH(AAeq)]     (uβy)=0βdu1ZTr[eβHeuHBeuH(AAeq)]=0βdueuHBeuH(AAeq)eq=0βduB(iu)(AAeq)eq     (euHを時間発展演算子とみなした)=0βduB(A(iu)Aeq)eq     (平衡状態の平均は時間に依存しない)=0βdu(BA(iu)eqBeqAeq)=0βduΔBΔA(iu)eq     (ref.  Cov(xy)=E[xy]E[x]E[y])\begin{aligned} \chi^{T}_{AB} &= \left. \frac{\partial}{\partial F_0} \frac{\text{Tr}\left[ e^{-\beta (H - F_0 B)} A \right]}{\text{Tr}\left[ e^{-\beta(H - F_0B)} \right]} \right|_{F_0 \to 0} \\ &= \left. \frac{1}{Z} \left\{ \frac{\partial}{\partial F_0} \text{Tr}\left[ e^{-\beta (H - F_0 B)} A \right] - \langle A \rangle_{eq} \frac{\partial}{\partial F_0} \text{Tr}\left[ e^{-\beta(H - F_0B)} \right] \right\} \right|_{F_0 \to 0} \\ &= \frac{1}{Z} \left\{ \text{Tr} \left[ \int^1_0 dy e^{-\beta H(1-y)} (\beta B) e^{-\beta H y}A \right] - \langle A \rangle_{eq} \text{Tr} \left[\int^1_0 dy e^{-\beta H(1-y)} (\beta B) e^{-\beta H y} \right] \right\} \\ &= \frac{1}{Z} \text{Tr} \left[\int^1_0 dy e^{-\beta H(1-y)} (\beta B) e^{-\beta H y} (A - \langle A \rangle_{eq}) \right] \\ &= \frac{1}{Z} \text{Tr} \left[\int^{\beta}_0 du e^{(u H-\beta H)} B e^{-u H} (A - \langle A \rangle_{eq}) \right] \ \ \ \ \ (u \coloneqq \beta y) \\ &= \int^{\beta}_0 du \frac{1}{Z} \text{Tr} \left[ e^{-\beta H} e^{u H} B e^{-u H} (A - \langle A \rangle_{eq}) \right] \\ &= \int^{\beta}_0 du \langle e^{u H} B e^{-u H} (A - \langle A \rangle_{eq}) \rangle_{eq} \\ &= \int^{\beta}_0 du \langle B_{(-i\hbar u)} (A - \langle A \rangle_{eq}) \rangle_{eq} \ \ \ \ \ \left( \because e^{u H}\text{を時間発展演算子とみなした} \right) \\ &= \int^{\beta}_0 du \langle B (A_{(i \hbar u)} - \langle A \rangle_{eq}) \rangle_{eq} \ \ \ \ \ \left( \because \text{平衡状態の平均は時間に依存しない} \right) \\ &= \int^{\beta}_0 du \left( \langle B A_{(i \hbar u)} \rangle_{eq} - \langle B \rangle_{eq} \langle A \rangle_{eq} \right) \\ &= \int^{\beta}_0 du \langle \Delta B \Delta A_{(i \hbar u)} \rangle_{eq} \ \ \ \ \ \left( \text{ref.} \ \ \text{Cov}(xy) = \text{E}[xy] - \text{E}[x]\text{E}[y] \right) \\ \end{aligned}

熱力学的応答関数と力学的応答関数の差 (2.35)\cdots(2.35)

limω0χAB(ω)=limω00dtϕAB(t)eiωt=limτ0τdtϕAB(t)=limτ0τdt0βduB˙(iu)A(t)eq     ((2.25))=limτ0τdt0βduddtBA(t+iu)eq     ((公式1))=0βdulimτ[BA(t+iu)eq]0τ=0βdu(BA(iu)eqlimτBA(τ+iu)eq)\begin{aligned} \lim_{\omega \to 0} \chi_{AB(\omega)} &= \lim_{\omega \to 0} \int^{\infty}_0 dt \phi_{AB(t)} e^{-i \omega t} \\ &= \lim_{\tau \to \infty} \int^{\tau}_0 dt \phi_{AB(t)} \\ &= \lim_{\tau \to \infty} \int^{\tau}_0 dt \int^{\beta}_0 du \langle \dot{B}_{(-i \hbar u)} A_{(t)} \rangle_{eq} \ \ \ \ \ (\because \text{(2.25)}) \\ &= - \lim_{\tau \to \infty} \int^{\tau}_0 dt \int^{\beta}_0 du \frac{d}{dt} \langle B A_{(t + i \hbar u)} \rangle_{eq} \ \ \ \ \ (\because (公式1)) \\ &= - \int^{\beta}_0 du \lim_{\tau \to \infty} \left[ \langle B A_{(t + i \hbar u)} \rangle_{eq} \right]^{\tau}_0 \\ &= \int^{\beta}_0 du \left( \langle B A_{(i \hbar u)} \rangle_{eq} - \lim_{\tau \to \infty} \langle B A_{(\tau + i \hbar u)} \rangle_{eq} \right) \end{aligned}

χABTχAB=0βduΔBΔA(iu)0βdu(BA(iu)eqlimτBA(τ+iu)eq)=0βdu(BA(iu)eqBeqAeq)0βdu(BA(iu)eqlimτBA(τ+iu)eq)=0βdu(limτBA(τ+iu)eqBeqAeq)=limτ0βdu(BA(τ+iu)eqBeqAeq)=limτ0βduΔBΔA(τ+iu)eq\begin{aligned} \chi^{T}_{AB} - \chi_{AB} &= \int^{\beta}_0 du \langle \Delta B \Delta A_{(i \hbar u)} \rangle - \int^{\beta}_0 du \left( \langle B A_{(i \hbar u)} \rangle_{eq} - \lim_{\tau \to \infty} \langle B A_{(\tau + i \hbar u)} \rangle_{eq} \right) \\ &= \int^{\beta}_0 du \left( \langle B A_{(i \hbar u)} \rangle_{eq} - \langle B \rangle_{eq} \langle A \rangle_{eq} \right) - \int^{\beta}_0 du \left( \langle B A_{(i \hbar u)} \rangle_{eq} - \lim_{\tau \to \infty} \langle B A_{(\tau + i \hbar u)} \rangle_{eq} \right) \\ &= \int^{\beta}_0 du \left( \lim_{\tau \to \infty} \langle B A_{(\tau + i \hbar u)} \rangle_{eq} - \langle B \rangle_{eq} \langle A \rangle_{eq} \right) \\ &= \lim_{\tau \to \infty} \int^{\beta}_0 du \left( \langle B A_{(\tau + i \hbar u)} \rangle_{eq} - \langle B \rangle_{eq} \langle A \rangle_{eq} \right) \\ &= \lim_{\tau \to \infty} \int^{\beta}_0 du \langle \Delta B \Delta A_{(\tau + i \hbar u)} \rangle_{eq} \end{aligned}

§ 2.2.2.2 揺動散逸定理 (P. 29)

時間反転

[A(t),B]eq=Tr[π[A(t),B]]=Tr[(Tπ[A(t),B]T)]=Tr[[B,A(t)]π]=Tr[[A(t),B]π]=[A(t),B]eq\begin{aligned} \langle [A_{(t)}, B] \rangle_{eq} &= \text{Tr} \left[ \pi [A_{(t)}, B] \right] \\ &= \text{Tr} \left[ \left( T \pi [A_{(t)}, B] T^{\dagger} \right)^{\dagger} \right] \\ &= \text{Tr} \left[ [B^{\dagger}, A^{\dagger}_{(-t)}] \pi^{\dagger} \right] \\ &= - \text{Tr} \left[ [A_{(-t)}, B] \pi \right] \\ &= - \langle [A_{(-t)}, B] \rangle_{eq} \\ \end{aligned}

[A(t),B]eq=[B(t),B]eq=[B,B(t)]eq=[B(t),B]eq=[A(t),B]eq\begin{aligned} \langle [A_{(t)}, B] \rangle_{eq} &= \langle [B_{(t)}, B] \rangle_{eq} \\ &= \langle [B, B_{(-t)}] \rangle_{eq} \\ &= - \langle [B_{(-t)}, B] \rangle_{eq} \\ &= - \langle [A_{(-t)}, B] \rangle_{eq} \\ \end{aligned}

[A(t),B]eq=[B˙(t),B]eq=ddt[B(t),B]eq=ddt[B,B(t)]eq=ddt[B(t),B]eq=[B˙(t),B]eq=[A(t),B]eq\begin{aligned} \langle [A_{(t)}, B] \rangle_{eq} &= \langle [\dot{B}_{(t)}, B] \rangle_{eq} \\ &= \frac{d}{dt} \langle [B_{(t)}, B] \rangle_{eq} \\ &= \frac{d}{dt} \langle [B, B_{(-t)}] \rangle_{eq} \\ &= - \frac{d}{dt} \langle [B_{(-t)}, B] \rangle_{eq} \\ &= \langle [\dot{B}_{(-t)}, B] \rangle_{eq} \\ &= \langle [A_{(-t)}, B] \rangle_{eq} \\ \end{aligned}

応答関数ϕAB\phi_{AB}

ϕAB(ω)=dti[A(t),B]eqeiωt=0dti[A(t),B]eq(eiωteiωt)=2i0dti[A(t),B]eqIm[eiωt]=2iIm[0dti[A(t),B]eqeiωt]=2iIm[χAB(ω)]=2χAB(ω)\begin{aligned} \phi_{AB(\omega)} &= \int^{\infty}_{\infty} dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} e^{i\omega t} \\ &= \int^{\infty}_0 dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} (e^{i \omega t} - e^{-i \omega t}) \\ &= 2i \int^{\infty}_0 dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} \text{Im}[e^{i \omega t}] \\ &= 2i \text{Im} \left[ \int^{\infty}_0 dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} e^{i \omega t} \right] \\ &= 2i \text{Im} [\chi_{AB(\omega)}] \\ &= 2 \chi'_{AB(\omega)} \end{aligned}

ϕAB(ω)=dti[A(t),B]eqeiωt=0dti[A(t),B]eq(eiωt+eiωt)=20dti[A(t),B]eqRe[eiωt]=2Re[0dti[A(t),B]eqeiωt]=2Re[χAB(ω)]=2χAB(ω)\begin{aligned} \phi_{AB(\omega)} &= \int^{\infty}_{\infty} dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} e^{i\omega t} \\ &= \int^{\infty}_0 dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} (e^{i \omega t} + e^{-i \omega t}) \\ &= 2 \int^{\infty}_0 dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} \text{Re}[e^{i \omega t}] \\ &= 2 \text{Re} \left[ \int^{\infty}_0 dt \frac{i}{\hbar} \langle [A_{(t)}, B] \rangle_{eq} e^{i \omega t} \right] \\ &= 2 \text{Re} [\chi_{AB(\omega)}] \\ &= 2 \chi''_{AB(\omega)} \end{aligned}

KMS 条件

AB(t+iβ)eq=Tr[πAeβHB(t)eβH]=1ZTr[eβHAeβHB(t)eβH]=1ZTr[AeβHB(t)eβHeβH]=1ZTr[AeβHB(t)]=1ZTr[eβHB(t)A]=B(t)Aeq\begin{aligned} \langle AB_{(t + i\hbar\beta)} \rangle_{eq} &= \text{Tr}[\pi A e^{-\beta H} B_{(t)} e^{\beta H}] \\ &= \frac{1}{Z} \text{Tr} [e^{-\beta H} A e^{-\beta H} B_{(t)} e^{\beta H}] \\ &= \frac{1}{Z} \text{Tr} [A e^{-\beta H} B_{(t)} e^{\beta H} e^{-\beta H}] \\ &= \frac{1}{Z} \text{Tr} [A e^{-\beta H} B_{(t)}] \\ &= \frac{1}{Z} \text{Tr} [e^{-\beta H} B_{(t)} A] \\ &= \langle B_{(t)} A \rangle_{eq} \end{aligned}

ΔBΔA(t+iβ)eq=ΔA(t)ΔBeq\langle \Delta B \Delta A_{(t + i\hbar\beta)} \rangle_{eq} = \langle \Delta A_{(t)} \Delta B \rangle_{eq}

CdzΔBΔA(z)eqeiωz=0dtΔBΔA(t)eqeiωt=dtΔBΔA(t+iβ)eqeiω(t+iβ)=eβωdtΔBΔA(t+iβ)eqeiωt=eβωdtΔA(t)ΔBeqeiωt\begin{aligned} \int_{C} dz \langle \Delta B \Delta A_{(z)} \rangle_{eq} e^{i \omega z} &= 0 \\ \therefore \int^{\infty}_{-\infty} dt \langle \Delta B \Delta A_{(t)} \rangle_{eq} e^{i \omega t} &= \int^{\infty}_{-\infty} dt \langle \Delta B \Delta A_{(t + i \hbar \beta)} \rangle_{eq} e^{i \omega (t + i \hbar \beta)} \\ &= e^{- \hbar \beta \omega}\int^{\infty}_{-\infty} dt \langle \Delta B \Delta A_{(t + i \hbar \beta)} \rangle_{eq} e^{i \omega t} \\ &= e^{- \hbar \beta \omega} \int^{\infty}_{-\infty} dt \langle \Delta A_{(t)} \Delta B \rangle_{eq} e^{i \omega t} \\ \end{aligned}

応答関数ϕAB\phi_{AB}

[A(t),B]=[ΔA(t)+Aeq,ΔB+Beq]=[ΔA(t),ΔB]\begin{aligned} [ A_{(t)}, B ] &= [ \Delta A_{(t)} + \langle A \rangle_{eq}, \Delta B + \langle B \rangle_{eq}] \\ &= [ \Delta A_{(t)}, \Delta B ] \end{aligned}

ϕAB=idt[A(t),B]eqeiωt=idt[ΔA(t),ΔB]eqeiωt=idt(ΔA(t)ΔBeqΔBΔA(t)eq)eiωt=idt(ΔA(t)ΔBeqeβωΔA(t)ΔBeq)eiωt=i(1eβω)dtΔA(t)ΔBeqeiωt\begin{aligned} \phi_{AB} &= \frac{i}{\hbar} \int^{\infty}_{-\infty} dt \langle [A_{(t)}, B] \rangle_{eq} e^{i \omega t} \\ &= \frac{i}{\hbar} \int^{\infty}_{- \infty} dt \langle [ \Delta A_{(t)}, \Delta B ] \rangle_{eq} e^{i \omega t} \\ &= \frac{i}{\hbar} \int^{\infty}_{- \infty} dt \left( \langle \Delta A_{(t)}\Delta B \rangle_{eq} - \langle \Delta B \Delta A_{(t)} \rangle_{eq} \right) e^{i \omega t} \\ &= \frac{i}{\hbar} \int^{\infty}_{- \infty} dt \left( \langle \Delta A_{(t)}\Delta B \rangle_{eq} - e^{-\hbar\beta\omega} \langle \Delta A_{(t)} \Delta B \rangle_{eq} \right) e^{i \omega t} \\ &= \frac{i}{\hbar} (1 - e^{-\hbar\beta\omega}) \int^{\infty}_{- \infty} dt \langle \Delta A_{(t)}\Delta B \rangle_{eq} e^{i \omega t} \\ \end{aligned}

ϕAB=i(1eβω)dtΔA(t)ΔBeqeiωt=(i1eβω)dteβωΔBΔA(t)eqeiωt=i(eβω1)dtΔBΔA(t)eqeiωt\begin{aligned} \phi_{AB} &= \frac{i}{\hbar} (1 - e^{-\hbar\beta\omega}) \int^{\infty}_{- \infty} dt \langle \Delta A_{(t)}\Delta B \rangle_{eq} e^{i \omega t} \\ &= (\frac{i}{\hbar} 1 - e^{-\hbar\beta\omega}) \int^{\infty}_{- \infty} dt e^{\hbar\beta\omega} \langle \Delta B \Delta A_{(t)} \rangle_{eq} e^{i \omega t} \\ &= \frac{i}{\hbar} (e^{\hbar\beta\omega} - 1) \int^{\infty}_{- \infty} dt \langle \Delta B \Delta A_{(t)} \rangle_{eq} e^{i \omega t} \\ \end{aligned}

CAB=dt12[ΔA(t)ΔBeq+ΔBΔA(t)eq]eiωt=12[dtΔA(t)ΔBeqeiωt+dtΔBΔA(t)eqeiωt]=12[1i(1eβω)ϕAB+1i(eβω1)ϕAB]=2ie12βω+e12βωe12βωe12βωϕAB=ω2iωcoth(βω2)ϕABϕAB=iω(ω2)coth(βω2)CAB\begin{aligned} C_{AB} &= \int^{\infty}_{-\infty} dt \frac{1}{2} \left[ \langle \Delta A_{(t)} \Delta B \rangle_{eq} + \langle \Delta B \Delta A_{(t)} \rangle_{eq} \right] e^{i\omega t} \\ &= \frac{1}{2} \left[ \int^{\infty}_{-\infty} dt \langle \Delta A_{(t)} \Delta B \rangle_{eq} e^{i\omega t} + \int^{\infty}_{-\infty} dt \langle \Delta B \Delta A_{(t)} \rangle_{eq} e^{i\omega t} \right] \\ &= \frac{1}{2} \left[ \frac{1}{\frac{i}{\hbar} (1 - e^{-\hbar\beta\omega})} \phi_{AB} + \frac{1}{\frac{i}{\hbar} (e^{\hbar\beta\omega} - 1)} \phi_{AB} \right] \\ &= \frac{\hbar}{2i} \frac{e^{\frac12 \hbar\beta\omega} + e^{-\frac12 \hbar\beta\omega}}{e^{\frac12 \hbar\beta\omega} - e^{-\frac12 \hbar\beta\omega}} \phi_{AB} \\ &= \frac{\frac{\hbar\omega}{2}}{i\omega} \coth{\left( \frac{\hbar\beta\omega}{2} \right)} \phi_{AB} \\ \therefore \phi_{AB} &= \frac{i\omega}{\left( \frac{\hbar\omega}{2} \right) \coth{\left( \frac{\beta\hbar\omega}{2} \right)}} C_{AB} \end{aligned}

散逸とゆらぎの関係

χAB=12iϕAB=1coth(βω2)CAB\begin{aligned} \chi''_{AB} &= \frac{1}{2i} \phi_{AB} \\ &= \frac{1}{\hbar \coth{\left( \frac{\beta\hbar\omega}{2} \right)}} C_{AB} \end{aligned}

CAB˙=dt12[ΔA(t)ΔB˙eq+ΔB˙ΔA(t)eq]eiωt=dt12(ddt[ΔA(t)ΔBeq+ΔBΔA(t)eq])eiωt     ((公式1))=dt12[ΔA(t)ΔBeq+ΔBΔA(t)eq]ddteiωt     (混合性)=iωdt12[ΔA(t)ΔBeq+ΔBΔA(t)eq]eiωt=iωCAB\begin{aligned} C_{A\dot{B}} &= \int^{\infty}_{-\infty} dt \frac12 \left[ \langle \Delta A_{(t)} \Delta \dot{B} \rangle_{eq} + \langle \Delta \dot{B} \Delta A_{(t)} \rangle_{eq} \right] e^{i\omega t} \\ &= - \int^{\infty}_{-\infty} dt \frac12 \left( \frac{d}{dt} \left[ \langle \Delta A_{(t)} \Delta B \rangle_{eq} + \langle \Delta B \Delta A_{(t)} \rangle_{eq} \right] \right) e^{i\omega t} \ \ \ \ \ (\because (\text{公式1})) \\ &= \int^{\infty}_{-\infty} dt \frac12 \left[ \langle \Delta A_{(t)} \Delta B \rangle_{eq} + \langle \Delta B \Delta A_{(t)} \rangle_{eq} \right] \frac{d}{dt} e^{i\omega t} \ \ \ \ \ (\because \text{混合性}) \\ &= i \omega \int^{\infty}_{-\infty} dt \frac12 \left[ \langle \Delta A_{(t)} \Delta B \rangle_{eq} + \langle \Delta B \Delta A_{(t)} \rangle_{eq} \right] e^{i\omega t} \\ &= i \omega C_{AB} \\ \end{aligned}

χAB=12ϕAB=iωωcoth(βω2)CAB=1ωcoth(βω2)CAB˙\begin{aligned} \chi'_{AB} &= \frac12 \phi_{AB} \\ &= \frac{i\omega}{\hbar\omega \coth{\left( \frac{\beta\hbar\omega}{2} \right)}} C_{AB} \\ &= \frac{1}{\hbar\omega \coth{\left( \frac{\beta\hbar\omega}{2} \right)}} C_{A\dot{B}} \\ \end{aligned}

タイプ 2: ω0\omega \to 0の極限

χAB=1ωcoth(βω2)CAB˙β2βω2βω2dtCAB˙(to)     (1coth(x)=tanh(x)=x+O(x3))=β2dtCAB˙(t)\begin{aligned} \chi'_{AB} &= \frac{1}{\hbar\omega \coth{\left( \frac{\beta\hbar\omega}{2} \right)}} C_{A\dot{B}} \\ &\to \frac{\frac{\beta}{2} \frac{\beta\hbar\omega}{2}}{\frac{\beta\hbar\omega}{2}} \int^{\infty}_{-\infty} dt C_{A\dot{B}(to)} \ \ \ \ \ \left( \because \frac{1}{\coth{(x)}} = \tanh{(x)} = x + \text{O}(x^3) \right) \\ &= \frac{\beta}{2} \int^{\infty}_{-\infty} dt C_{A\dot{B}(t)} \end{aligned}

2.2.3 分布応答の線形応答と揺動散逸定理 (P. 31)

Htot=Hs+kHk+HksH_{tot} = H_s + \sum_{k} H_k + H_{ks}

分布関数 ρ0\rho_0

σueu(G+X)σˉueuGσu\begin{aligned} \sigma_u &\coloneqq e^{-u ( G + X )} \\ \bar{\sigma}_u &\coloneqq e^{uG}\sigma_u \end{aligned}

uσˉu=euGGσueuG(G+X)σu=euGXσu=euGXeuGσˉu=X(iu)σˉuσˉβ=e0βduX(iu)σβ=euGe0βduX(iu)\begin{aligned} \frac{\partial}{\partial u} \bar{\sigma}_u &= e^{uG} G \sigma_u - e^{uG} (G + X) \sigma_u \\ &= - e^{uG} X \sigma_u \\ &= - e^{uG} X e^{-uG} \bar{\sigma}_u \\ &= - X_{(-i\hbar u)} \bar{\sigma}_u \\ \therefore \bar{\sigma}_\beta &= e^{-\int^{\beta}_0 du X_{(-i \hbar u)}} \\ \therefore \sigma_\beta &= e^{-uG} e^{-\int^{\beta}_0 du X_{(-i \hbar u)}} \\ \end{aligned}

σβ=eβGe0βduX(iu)=eβG(10βduX(iu))+O(X2)σβ=e0βduX(iu)euG     (σβはエルミート)=(10βduX(iu))eβG+O(X2)σβeβG0βdu12[eβGX(iu)+X(iu)eβG]\begin{aligned} \sigma_\beta &= e^{-\beta G} e^{-\int^{\beta}_0 du X_{(-i \hbar u)}} \\ &= e^{-\beta G} \left( 1 - \int^{\beta}_0 du X_{(-i \hbar u)} \right) + \text{O}_{(X^2)} \\ \sigma_\beta &= e^{-\int^{\beta}_0 du X_{(i \hbar u)}} e^{-uG} \ \ \ \ \ (\because \sigma_\beta \text{はエルミート}) \\ &= \left( 1 - \int^{\beta}_0 du X_{(i \hbar u)} \right) e^{-\beta G} + \text{O}_{(X^2)} \\ \therefore \sigma_\beta &\simeq e^{-\beta G} - \int^{\beta}_0 du \frac12 \left[ e^{-\beta G} X_{(-i \hbar u)} + X_{(i \hbar u)} e^{-\beta G} \right] \end{aligned}

Tr[eu(G+X)]Tr[eβG0βdu12[eβGX(iu)+X(iu)eβG]]=Tr[eβG]0βdu12(Tr[eβGX(iu)]+Tr[X(iu)eβG])=Z0βdu12(ZX(iu)eq+ZX(iu)eq)=Z(10βdu12(X(iu)eq+X(iu)eq))\begin{aligned} \text{Tr} \left[ e^{-u (G+X)} \right] &\simeq \text{Tr}\left[ e^{-\beta G} - \int^{\beta}_0 du \frac12 \left[ e^{-\beta G} X_{(-i \hbar u)} + X_{(i \hbar u)} e^{-\beta G} \right] \right] \\ &= \text{Tr}\left[ e^{-\beta G} \right] - \int^{\beta}_0 du \frac12 \left( \text{Tr}\left[ e^{-\beta G} X_{(-i \hbar u)} \right] + \text{Tr}\left[ X_{(i \hbar u)} e^{-\beta G} \right] \right) \\ &= Z - \int^{\beta}_0 du \frac{1}{2} ( Z \langle X_{(-i \hbar u)} \rangle_{eq} + Z \langle X_{(i \hbar u)} \rangle_{eq}) \\ &= Z \left( 1 - \int^{\beta}_0 du \frac{1}{2} ( \langle X_{(-i \hbar u)} \rangle_{eq} + \langle X_{(i \hbar u)} \rangle_{eq}) \right) \end{aligned}

11x=1+x+O(x2)\frac{1}{1-x} = 1 + x + \text{O}_{(x^2)}

ρ0=eβ(G+X)Treβ(G+X)(eβG0βdu12[eβGX(iu)+X(iu)eβG]) 1Z(1+0βdu12(X(iu)eq+X(iu)eq))=(π0βdu12[πX(iu)+X(iu)π])(1+0βdu12(X(iu)eq+X(iu)eq))π0βdu12[πX(iu)+X(iu)π]+0βdu12(πX(iu)eq+X(iu)eqπ)=π0βdu12[πΔX(iu)+ΔX(iu)π]\begin{aligned} \rho_0 &= \frac{e^{-\beta (G + X)}}{\text{Tr} e^{-\beta (G + X)}} \\ &\simeq \left( e^{-\beta G} - \int^{\beta}_0 du \frac12 \left[ e^{-\beta G} X_{(-i \hbar u)} + X_{(i \hbar u)} e^{-\beta G} \right] \right) \frac{1}{Z} \left( 1 + \int^{\beta}_0 du \frac{1}{2} ( \langle X_{(-i \hbar u)} \rangle_{eq} + \langle X_{(i \hbar u)} \rangle_{eq}) \right) \\ &= \left( \pi - \int^{\beta}_0 du \frac12 \left[ \pi X_{(-i \hbar u)} + X_{(i \hbar u)} \pi \right] \right) \left( 1 + \int^{\beta}_0 du \frac{1}{2} ( \langle X_{(-i \hbar u)} \rangle_{eq} + \langle X_{(i \hbar u)} \rangle_{eq}) \right) \\ &\simeq \pi - \int^{\beta}_0 du \frac12 \left[ \pi X_{(-i \hbar u)} + X_{(i \hbar u)} \pi \right] + \int^{\beta}_0 du \frac{1}{2} ( \pi \langle X_{(-i \hbar u)} \rangle_{eq} + \langle X_{(i \hbar u)} \rangle_{eq} \pi) \\ &= \pi - \int^{\beta}_0 du \frac12 \left[ \pi \Delta X_{(-i \hbar u)} + \Delta X_{(i \hbar u)} \pi \right] \end{aligned}

演算子の時間発展 (2.55)\cdots (2.55)

A(τ)eq=Tr[ρ0UAU]=Tr[AUρ0U]=Tr[AUπU]0βdu12(Tr[AUπΔX(iu)U]+Tr[AUΔX(iu)πU])=Tr[πA]0βdu12(Tr[πΔX(iuτ)A]+Tr[πAΔX(iuτ)])=Aeq0βdu12[ΔX(iuτ)Aeq+AΔX(iuτ)eq]=Aeq+0βduτ0dt12[ΔX˙(iu+t)eq+AΔX˙(iu+t)eq]0βdu12(ΔX(iu)Aeq+AΔX(iu)eq)\begin{aligned} \langle A_{(\tau)} \rangle_{eq} &= \text{Tr}[ \rho_0 U^{\dagger} A U] \\ &= \text{Tr}[A U \rho_0 U^{\dagger}] \\ &= \text{Tr}[AU\pi U^{\dagger}] - \int^{\beta}_0 du \frac12 \left( \text{Tr}[AU\pi\Delta X_{(-i \hbar u)}U^{\dagger}] + \text{Tr}[AU\Delta X_{(i \hbar u)} \pi U^{\dagger}] \right) \\ &= \text{Tr}[\pi A] - \int^{\beta}_0 du \frac12 \left( \text{Tr}[\pi\Delta X_{(-i \hbar u - \tau)} A] + \text{Tr}[\pi A\Delta X_{(i \hbar u - \tau)}] \right) \\ &= \langle A \rangle_{eq} - \int^{\beta}_0 du \frac12 \left[ \langle \Delta X_{(-i \hbar u - \tau)} A \rangle_{eq} + \langle A\Delta X_{(i \hbar u - \tau)} \rangle_{eq} \right] \\ &= \langle A \rangle_{eq} + \int^{\beta}_0 du \int^0_{-\tau} dt \frac12 \left[ \langle \Delta \dot{X}_{(-i \hbar u + t)} \rangle_{eq} + \langle A \Delta \dot{X}_{(i \hbar u + t)} \rangle_{eq} \right] - \int^{\beta}_0 du \frac12 \left( \langle \Delta X_{(-i \hbar u)} A \rangle_{eq} + \langle A\Delta X_{(i \hbar u)} \rangle_{eq} \right) \end{aligned}